If $\frac{|z-a|}{|z-b|}=c$ for a constant $c$ not equal to $1$, then prove that the expression is a circle

Question

I saw a similar question which asked to show that the locus traced out by $arg(\frac{z-a}{z-b})=c$ is a circle. I know how to prove that but what is the connection between the $2$ questions. Can I describe the equation $\frac{z-a}{z-b}=c$ by its argument?

Also, how do I find the radius and centre of the equation?

Thanks

Answer 1

Let $$\frac{|z-a|}{|z-b|}=c.$$

Let $z=x+iy$ and $a=a_1+ia_2$, $b=b_1+ib_2.$

We have $$(x-a_1)^2+(y-a_2)^2= c^2(x-b_1)^2+c^2(y-b_2)^2$$

$$ (1-c^2)x^2 + ( 1-c^2)y^2 -2(a_1-c^2 b_1)x-2(a_2-c^2 b_2)y+a_1^2+b_2^2-c^2b_1^2-c^2b_2^2=0$$

Which is the equation of a circle if the constant $ a_1^2+b_2^2-c^2 b_1^2-c^2 b_2^2$ is negative.

The center and the radius could easily be found.